Recent content by clamtrox

Since T>>m, the distribution has most of its weight on values E>>m. Therefore E^2 - m^2 ~ E^2 as far as that integral is concerned.

I don't have Carrolls book, but I think he explains this very well in his lecture notes. See page 140, equation (5.43) and the text around that equation. The same property holds for any geodesic. The affine parameter is only defined up to some scaling factor \lambda \rightarrow a \lambda + b...

1. Let i label the match, and j label the game inside a match, and i=1...N and j=1...ki where ki is the number of games in given match. Also let dij be the result of the game j in match i (1 for a win and 0 for a loss). Then  p = \frac{1}{N} \sum_{i=1}^{N} \frac{1}{k_i} \sum_{j=1}^{k_i} d_{ij}...

Can you explain your notation a little more? What are p and p'?

Method 1 gives equal weight to every game, whereas method 2 gives equal weight to every point. I don't see why you would want to give equal weight to every game, ie. giving more weight to points scored in fast games (with only 3 or 4 points).

It's correct. But it's not the best way to estimate the probability, as you only have 10 data points. If you are able to use the individual rounds, you have much more data, and therefore you can determine the probability more accurately. This is what HallsofIvy was showing you.

You can't really estimate this without having a model of how the points are correlated. Assuming that match scores are uncorrelated (so that result of a match is not affected by previous results), then you can just calculate the average of matches won, and also estimate the standard deviation...

It's very appropriate to use the "direct" measurement value, for example http://arxiv.org/abs/1103.2976. This is based on distance laddering up to supernovae, and is fairly independent of any cosmological assumptions. I wouldn't use the Planck value though, as that has much more modelling...

It depends on the definition. You can see that Wikipedia defines the coupling constant as \alpha_G = m_e / m_p , whereas normally you would call 1/m_p as the coupling constant, as it's what's in front of the interaction term in the Lagrangian.

\sum_{n=1}^\infty x^n = x \sum_{n=0}^\infty x^n

Perhaps you don't need to calculate the potential at all. Seems like you get work directly from the force, according to your formula. The path parametrization depends on the path of course. It seems a path is not specified in the problem. Why is it not needed to solve the problem?

I can't follow at all what you have done, so I can't comment on it. Perhaps the best approach would be to use the commonly known result for geometric series: \sum_{n=0}^\infty x^n = \frac{1}{1-x} for |x| < 1

The charge distribution is just \rho = -e \sum |\Psi_{nlm}|^2 where the sum is over all the electrons you're interested in. In this case, they are the electrons with n=3, l=2 so you should sum is over m and ms, for all values possible for this shell.

Yap that looks correct

So m is the eigenvalue corresponding to operator \hat{L}_z , and ms corresponds to \hat{S}_z . You basically got it, now you just need to figure out which values can m and ms take together, and how many combinations there are.